2021 Fall AMC 12A Problems/Problem 11
Contents
Problem
Consider two concentric circles of radius and The larger circle has a chord, half of which lies inside the smaller circle. What is the length of the chord in the larger circle?
Solution 1 (Pythagorean Theorem)
Label the center of both circles . Label the chord in the larger circle as , where and are on the larger circle and and are on the smaller circle. Construct the radius perpendicular to the chord and label their intersection as . Because a radius that is perpendicular to a chord bisects the chord, is the midpoint of the chord.
Construct segments and . These are radii with lengths 17 and 19 respectively.
Then, use the Pythagorean Theorem. In , we have
In , we have
Equating these two expressions, we get and .
~eisthefifthletter and Steven Chen
Solution 2 (Power of a Point)
Draw the diameter perpendicular to the chord. Notice that by symmetry this diameter bisects the chord.
Call the intersection between that diameter and the chord . In the smaller circle, let the shorter piece of the diameter cut by the chord be , making the longer piece In that same circle, let the be the length of the portion of the chord that is cut by the diameter.
The radius of the larger circle is more than the radius of the small circle. So, in the larger circle, the shorter piece of the diameter cut by the chord is of length and the longer piece is As given in the problem, the bisected length of the chord in the larger circle is twice as much, so it is of length . By Power of a Point, we can construct a system of equations Expanding both equations, we get in which the and terms magically cancel when we subtract the first equation from the second equation. Thus, now we have
-fidgetboss_4000
Solution 3 (Coordinate Geometry)
Represent the circles as , and . Solving for in these equations we obtain and . Because half of the chord is in the smaller circle, the larger circle should have an value that is twice as big as the smaller circle's value. We set up the equation: Substituting into , we obtain . However, this is only half of the chord length, so we must double it to obtain
-helpmebro
Video Solution by TheBeautyofMath
~IceMatrix
Video Solution (Logic and Geometry)
~Education, the Study of Everything
See Also
2021 Fall AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 10 |
Followed by Problem 12 |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
All AMC 12 Problems and Solutions |
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